Course Plan ICT sem-1 BEC
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Transcript of Course Plan ICT sem-1 BEC
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Course plan
Subject : Basic Electric circuits(ICT 102) Teaching plan per week 3 lecture-periods Most lectures in tutorial mode Must bring calculator for every lecture Two weeks devoted to project
presentations
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Text Book
INTRODUCTION TO ELECTRIC CIRCUITS 9th Edition Richard C. Dorf and James A. Svoboda Chapters 1 to 10 complete Parts of 11th chapter
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Syllabus
Electric circuit variables; Circuit elements : Resistive circuits: Methods of analysis of resistive circuits: Circuit theorems : The operational amplifier : Energy storage elements : Response of RLC circuits : Sinusoidal steady state. Power calculations.
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Evaluation
Attendance 5% less than 50 % : 0 : 50-59 ; 1: 60-69 ;2 70-79 ; 3: 80-89;4 : 90-100 : 5 Quizzes 20 % Two Quizzes Mid semester test 30% End semester test 30% Project 15 %
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Project
Electrical wiring design (Home, theater,malletc) 8 Students in a team Sequence of work for project. Team building workshop A lecture on the project. Team compositions (1) Theory (2) Practice (3) Communication (4) Social skills
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Project
Presentations as and when declared. Preparing the layout-design Report writing Presentation of final design & Submission of report
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Teams for lab project
For lab two divisions of around 60 each Team of 4 for lab project Team of 2 for lab experiments
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Chapter 1
Units Electric circuit variables Current Voltage Power Energy Problem solving Design and analysis
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International system of units (SI units)
SI base units
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Standards of units of measurements
Measurements done at different places and times should be comparable
Standards are needed Early standards were physical objects that
defined the unit as one of the physical properties
Standard bars for kg and meter and a standard candle
Properties change with time Atomic standards are the best
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Derived Units in SI
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Examples
1 MHz = ? Hz 106 Hz 1 mF = ? F 10-6 Farad 1 KW = ? W 10 3 W
1 mH = ? H 10-3 H 1 volt = ? mV 106 mV 1 A = ? mA 103 mA
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Example 1.3-1 SI units
A mass of 150 grams experiences a force of 100 newtons. Find the energy or work expended if the mass moves 10 cms. Also find the power if the mass completes its move in 1 millisecond
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Solution to example 1.3-1
Energy = Force x distance = 100 x 0.1 = 10 joulesNote that the distance is in metersPower = Energy/Time period = 10/10-3
= 104 joules/second = 104 watts = 10 kwatts
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Exercise 1.3-1
is largest ?
i1 = 45 mA : i2 = .03x103 = 30 mA i3 = 25x10-4x 106 = 2500 mA
i3 is the largest
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Electrical circuit
Electrical elements connected together Used in Generation, transmission and
consumption of electric power and energy
Storage , transmission and processing of information
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Charge on electron
- 1.602x10-19 coulombs -1 coulomb is the charge on 6.24x1018
electrons Electrons flow from –ve terminal of
battery to +ve terminal Conventionally direction of current is
taken from +ve to –ve terminal
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Charge and current calculations
calculatedbecanqgivenisiifSo
0tatchargetheisq(0)Where
qididq
havealsoWe
calculatedbecanigivenischargeIfdt
dqi
t t
)0(
:
0
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Nature of current
Direct current ; this is a current of constant magnitude
Battery connected to resistors Time varying current Circuit connected to mains Charging of a capacitor Discharging of a capacitor
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A direct current of magnitude I
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Examples of time varying current
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Exercise 1.2-1
Find the charge that has entered the element by time t when
i = 8t2-4t A Assume q(t) = 0 for t < 0
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Solution to exercise 1.2-1
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Exercise 1.2-2
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Solution to Exercise 1.2-2
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Voltage
In a battery electrons move from –ve terminal to +ve terminal.
To maintain the +ve charge on the terminal the electrons reaching it should be removed
This is done by chemical reaction in the battery
To take an electron away from the +ve terminal some work has to be done.
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Analogy
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Work done/Unit charge
Voltage is defined as the work done to move a +ve charge of 1 coulomb from negative terminal of battery to +ve terminal
Unit of energy (Work) is joule So voltage is joule/coulomb General equation for voltage is
dq
dwv
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Power and Energy
A light bulb absorbs energy from source The rate of absorbing the energy is
called power
ondsintimeistand
joulesinenergyiswwattsinpowerispWheredt
dwp
sec
,
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Significance of power and energy A light bulb is specified in terms of its
power A 300 watt bulb gives more light than
100 watt bulb Energy absorption will depend on power
and time both as it is obtained by integrating power over the given time period
Your electric bill depends on the energy consumed.
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Power in terms of voltage and current
ivdt
dq
dq
dw
dt
dwp ..
Power absorption and Power Supplied
When current enters the + ve terminal, power is absorbed by the element
When the current leaves the + ve terminal the element supplies power
Power absorbed = - power supplied
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Exercise 1.5-1
Figure E 1.5-1 shows four circuit elements identified by the letters A, B, C, and D. (a) Which of the devices supply 12 W? (b) Which of the devices absorb 12 W? (c) What is the value of the power received by device B? (d) What is the value of the power delivered by device B? (e) What is the value of the power delivered by device D?
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Problem types
Two types Problems with unique solutions Problems with more than one solution
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Example
Problem with unique solution A battery across a bulb Voltage is known Wattage of bulb known Find the resistance W = V2/R ; R = V2/W
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Problem with non-unique solution A battery across a bulb Wattage rating is specified Find out the battery voltage and the
resistance Now we have a equation W = V2/R W is known but V and R are unknown One equation but two unknowns Assume one and calculate the other Many pairs of V and R will satisfy the
equation
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Analysis problems
Sources and circuit is known Find the voltage or current through a
given element Unique solution
A simple problem in analysis
V1 = 10 voltsR1 = 10 kohmsR2 = 10 kohmsFind V2V2 = 5 volts
V2 = V1.R2/(R1+R2)
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A more general problem
More complicated circuit Many simultaneous equations Equations solved to get the required
voltage or current
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Design
One example Input and output is known Find the circuit
V1 = 1 voltV2 = 0.5 voltMany solutionsR1=R2= 1kohmR1=R2=10kohmR1 = R2 = 100 kohms
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A more involved design
Output is often user defined Audio amplifier for PA system P0 = 30 watts Good quality Designer has to chose the Gain Frequency response IC or transistor
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The problem solving method
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Design cycle
1 Get specifications 2 Design 3 Analyze 4 Are specs satisfied? No yes 5 Implement 6 Measure 7 Are specs satisfied? No 8 Stop
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P1.2-1
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Solution to P1.2-1
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P1.2-2
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Solution to P1.2-2
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P1.2-4
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Solution to P1.2-4
P 1.2-5
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Solution to 1.2-5
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P 1.2-6
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Solution to 1.2-6
P 1.3-4
Solution to P 1.3-4
P1.5-2
Solution to P1.5-2
P 1.5-4
Solution P 1.5-4
Solution to P1.5-4 Continued
P1.7-1
Solution to 1.7-1